$\dfrac{dy}{dt}=y$, and $y=1$ when $t=4$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $y=4e^{t-1}$ (Choice B) B $y=e^{t-4}$ (Choice C) C $y=e^{4t}$ (Choice D) D $y=4e^{t}$
The general solution of equations of the form $\dfrac{dy}{dt}=ky$ is $y=C\cdot e^{kt}$ for some constant $C$. This can be found using separation of variables. In our case, $k=1$, so $y=C\cdot e^{t}$. Let's use the fact that $y=1$ when $t=4$ to find $C$ : $\begin{aligned} y&=C\cdot e^{t} \\\\ 1&=C\cdot e^{4} \gray{\text{Plug }t=4\text{ and }y=1} \\\\ e^{-4}&=C \end{aligned}$ In conclusion, $y=e^{t-4}$.